Question

Помогите решить интеграл

Answer 1

$\int\frac{dx}{(8-3x)(4-5x)}=\frac{1}{28}\int\frac{d(8-3x)}{8-3x}-\frac{1}{28}\int\frac{d(4-5x)}{4-5x}=\\=\frac{1}{28}ln|8-3x|-\frac{1}{28}ln|4-5x|=\frac{1}{28}ln|\frac{8-3x}{4-5x}|$
$\frac{1}{(8-3x)(4-5x)}=\frac{A}{8-3x}+\frac{B}{4-5x}=\frac{-\frac{3}{28}}{8-3x}+\frac{\frac{5}{28}}{4-5x}\\1=A(4-5x)+B(8-3x)\\x^0|1=4A+8B\\x^1|0=-5A-3B\\\begin{cases}4A+8B=1|*5\\-5A-3B=0|*4\end{cases}=\ \textgreater \ \begin{cases}20A+40B=5\\+\\-20A-12B=0\end{cases}\\28B=5\\B=\frac{5}{28}\\4A+\frac{40}{28}=1\\4A=-\frac{12}{28}\\A=-\frac{3}{28}$